16.28 Envelopes and volumes of singularities168D3DF32068

Proof. The solutions exist by theorem 14.7.2. Choose $A\geq 0$ with $\mathrm{dd}^{\mathrm{c}}f\leq A\omega $. At a maximum point of $u_\beta -f$, the second derivative test gives

Evaluating the equation at this point, we obtain

Thus $u_\beta -C_ f/\beta $ is a competitor for $P_\omega (f)$, and $u_\beta \leq p+C_ f/\beta $.

For the lower bound, note that $p$ is continuous by proposition 5.40.3, and choose smooth strictly $\omega $-plurisubharmonic functions $v_ j\downarrow p$ by theorem 5.32.1; by Dini's theorem the convergence is uniform. For any $\delta >0$, choose $j$ so that

Then $v_ j-\delta \leq f$, since $p\leq f$. Since $\omega _{v_ j}>0$ and $X$ is compact, there is $c_ j>0$ such that $\omega _{v_ j}^ n\geq c_ j\omega ^ n$. Put $C_ j=\max (0,-\log c_ j)$. The function

satisfies

Exponential comparison, corollary 14.19.12 with $\lambda =\beta $, gives $w_{j,\beta }\leq u_\beta $. Consequently

Letting first $\beta \to \infty $ with $j$ fixed, so that no lower bound for $c_ j$ uniform in $j$ is needed, and then $\delta \downarrow 0$, we obtain uniform convergence. Bounded Bedford–Taylor continuity then gives weak convergence of the measures.

On a compact subset of the open set $\{ p<f\} $, uniform convergence gives $u_\beta -f\leq -\epsilon $ for some $\epsilon >0$ and all large $\beta $. There the measures $\omega _{u_\beta }^ n$ have densities at most $\mathrm{e}^{-\beta \epsilon }$ with respect to $\omega ^ n$. Testing against continuous functions with compact support in $\{ p<f\} $ and passing to the limit gives $\omega _ p^ n=0$ on this open set. $\square$

Proof. Suppose first that $f$ is continuous. Choose smooth functions $f_ j\to f$ uniformly. The Lipschitz bound for envelopes in proposition 5.40.3 gives $p_ j=P_\omega (f_ j)\to p=P_\omega (f)$ uniformly. Every compact subset of the open set $\{ p<f\} $ has a neighborhood on which $p_ j<f_ j$ for all large $j$. The preceding proposition and convergence of bounded products then give $\omega _ p^ n=0$ on $\{ p<f\} $.

For lower semicontinuous $f$, choose continuous functions $f_ j\uparrow f$ with common upper and lower bounds. For example, for a fixed distance $d$ inducing the topology of $X$, one may take

These functions are Lipschitz and increase to $f$: if convergence failed at $x$, approximate minimizers $y$ would approach $x$, contradicting lower semicontinuity of $f$ at $x$.

The functions $p_ j=P_\omega (f_ j)$ increase and stay bounded, so the regularized limit $p_\infty =(\lim _ jp_ j)^*$ is a bounded $\omega $-plurisubharmonic function. It agrees with $\lim _ jp_ j$ outside a pluripolar set, and since each $p_ j$ lies below $f_ j\leq f$ outside a pluripolar set, we obtain $p_\infty \leq f$ quasi-everywhere. Thus $p_\infty \leq P_\omega (f)$.

For fixed $k$, the set $\{ p_\infty <f_ k\} $ is open, since $p_\infty $ is upper semicontinuous and $f_ k$ is continuous. If $j\geq k$, then $p_ j\leq p_\infty <f_ k\leq f_ j$ there, so the continuous case gives $\omega _{p_ j}^ n=0$ on that open set. Bedford–Taylor continuity along increasing sequences applies after adding a fixed smooth local function whose Hessian dominates $\omega $ and expanding with respect to the resulting smooth form. Together with the lower semicontinuity of the mass of open sets under weak convergence, it yields

Taking the union over $k$ gives zero measure on $\{ p_\infty <f\} $, hence also on $\{ p_\infty <P_\omega (f)\} $ outside a pluripolar set, since $P_\omega (f)\leq f$ quasi-everywhere. Bounded measures of this kind do not charge pluripolar sets, so the domination principle, theorem 14.19.10, gives $p_\infty \geq P_\omega (f)$. The two functions are therefore equal, and the assertion follows. $\square$

Proof. First suppose that $\theta =\omega $ is Hermitian. Choose bounded lower semicontinuous functions $f_ j\downarrow f$ quasi-everywhere, with common bounds, as in lemma 5.29.15. The bounded functions $p_ j=P_\omega (f_ j)$ decrease to $p$. Indeed, their limit is a competitor for $p$, since it lies below $f$ quasi-everywhere, and the reverse inequality follows from monotonicity.

For fixed $k$, put $H_ k=\min \{ 1,(f-p_ k)_+\} $, a bounded quasi-continuous function. Where $H_ k>0$ and $j\geq k$, we have $p_ j\leq p_ k<f\leq f_ j$ outside a pluripolar set. The lower semicontinuous case therefore gives

The convergence of bounded products against quasi-continuous weights in section 5.29 gives $\int _ XH_ k\, \omega _ p^ n=0$. Since the sets $\{ p_ k<f\} $ increase to $\{ p<f\} $, this proves the assertion for Hermitian $\theta $.

For general semipositive $\theta $, set $\theta _ j=\theta +j^{-1}\omega $. The functions $p_ j=P_{\theta _ j}(f)$ have the common bounds $\inf _ Xf\leq p_ j\leq \sup _ Xf$ and decrease. Their limit is $\theta $-plurisubharmonic by distributional convergence and lies below $f$ quasi-everywhere, so it is at most $p=P_\theta (f)$; since $p\leq p_ j$ for all $j$, it equals $p$. For each fixed $k$, the same coefficient $H_ k=\min \{ 1,(f-p_ k)_+\} $ gives

All $p_ j$ are bounded potentials for the fixed Hermitian form $\theta +\omega $. Expanding the products with background $\theta _ j$ with respect to this fixed background, we may use convergence of bounded products and uniform capacity bounds to pass the last identity to the limit, with the quasi-continuous weight $H_ k$; this gives $\int _ XH_ k\, \mathrm{MA}_\theta (p)=0$. Exhausting $\{ p<f\} $ as before completes the proof. $\square$

Proof. The local envelope theorem, applied to a countable family realizing the supremum (Choquet's lemma), gives $p\leq f$ outside a pluripolar set. Choose $M>\sup _ Xp$ and replace $f$ by $\min (f,M)$. This does not change the envelope, since $p$ itself is a competitor for the smaller obstacle. Since $p<M$, the contact set is also unchanged, so it suffices to prove the theorem for the smaller obstacle; that is, we may assume that $f$ is bounded above.

Set $f_ j=\max (f,-j)$ and $p_ j=P_\theta (f_ j)$. The constants $-j$ are competitors, so $p_ j$ is bounded. Theorem 16.28.4 gives $p_ j\leq f_ j$ quasi-everywhere and

The sequence $p_ j$ decreases and stays above $p$. Its limit is therefore $\theta $-plurisubharmonic, it lies below $f$ quasi-everywhere, and so it is a competitor for $p$. Hence $p_ j\downarrow p$.

Fix an integer $k\geq 1$ and $C>0$, and define the bounded nonnegative quasi-continuous function

Its values on pluripolar sets are immaterial. Where this coefficient is positive, $p_ k<f$ and $p>-C$. For $j\geq k$, we then have $p_ j\leq p_ k<f\leq f_ j$ and $p_ j\geq p>-C$ there. Plurifine locality and theorem 16.28.4 therefore imply

The truncated potentials are uniformly bounded and decrease to $\max (p,-C)$, so their measures converge weakly by bounded Bedford–Taylor continuity. The uniform capacity bounds and the weighted convergence in section 5.29 also allow testing against $H_{k,C}$. Hence the last identity holds with $p$ in place of $p_ j$. By plurifine locality, this says that the extended measure of $p$ vanishes on $\{ p_ k<f\} \cap \{ p>-C\} $. Off the pole set of $p$, these sets increase to $\{ p<f\} $ as first $k$ and then $C$ tend to infinity. Continuity of positive measures along increasing sequences proves the assertion, even if the total extended mass is infinite. $\square$

Proof. The envelope lies below both obstacles quasi-everywhere, hence everywhere, since the obstacles are upper semicontinuous. By the contact theorem (theorem 16.28.5 or theorem 15.8.6), its measure is carried by $\{ p=\min (u,v)\} $. Partition this set into $\{ p=u<v\} $ and $\{ p=v\} $, and apply proposition 13.14.6 on each part. All measures vanish on the pluripolar pole sets, which can therefore be discarded throughout. $\square$

Proof. Since $\varphi =P_\theta [\varphi ]\leq 0$, the function $\varphi $ is a competitor for $P_\theta (\varphi +1,0)$. This rooftop is also a competitor for $P_\theta [\varphi ]=\varphi $, being at most $\varphi +1$ and at most zero, so it equals $\varphi $. Now apply proposition 16.28.6 with $u=\varphi +1$ and $v=0$. The contact set $\{ \varphi =\varphi +1<0\} $ is contained in the pole set of $\varphi $, which the extended measure does not charge. Only contact with zero contributes, and this gives the asserted bound. $\square$

Proof. Use homogeneous coordinates $[Z_0:\cdots :Z_ n]$ and the affine chart $Z_0\ne 0$. On a chart $Z_ i\ne 0$ meeting $H=\{ Z_0=0\} $, the local plurisubharmonic representative of $p=u-\delta \chi /2$ is

This function is locally bounded above near $H$, by the growth bound on $u$, because the norm of the affine coordinates is at most a constant, depending on the chart, divided by $|Z_0/Z_ i|$. By lemma 5.24.7, it therefore extends canonically across $H$. Since the local potentials of $\theta _\delta $ differ by smooth functions on overlaps, the extensions agree there.

Conversely, every $p\in \mathrm{PSH}(\mathbb P^ n,\theta _\delta )$ is bounded above, since $\mathbb P^ n$ is compact. The function $u=p+\delta \chi /2$ is plurisubharmonic on the affine chart, and

gives the required growth bound. Neither function is identically $-\infty $ on the affine chart. Uniqueness of the canonical extension shows that the two maps are inverse to each other. $\square$

Proof. By lemma 16.28.9, the map $u\mapsto u-\delta \chi /2$ is a bijection between the two families of competitors, which proves the identity for $V_\delta $. Constants below $\inf _ K(\psi -\delta \chi /2)$ are competitors for $G_\delta $.

Since all competitors for $G_\delta $ lie below the constant $\sup _ K(\psi -\delta \chi /2)$ on $K$, lemma 5.40.1 gives a common upper bound for them. Together with a constant competitor, the upper-envelope theorem shows that $G_\delta ^*$ is bounded and $\theta _\delta $-plurisubharmonic. Addition of the smooth function $\delta \chi /2$ in (16.28.11) commutes with regularization, and the two-sided bounds for $G_\delta ^*$ give $F_\delta \in \mathcal L_\delta ^+$.

Locally, adding a potential of $\theta _\delta $ makes the competitors plurisubharmonic. The upper-envelope theorem recalled in section 5.21 shows that regularization changes their supremum only on a pluripolar set. Thus $G_\delta =G_\delta ^*$ almost everywhere, and adding $\delta \chi /2$ gives the same assertion for $V_\delta $. $\square$

Proof. Put $E_ j=E\cap \overline B(0,j)$. Each $E_ j$ is pluripolar as a subset of $\mathbb P^ n$: it is locally pluripolar near every affine point, and a neighborhood of the hyperplane at infinity does not meet this bounded set. By Josefson's theorem on compact manifolds, [Theorem 1.1.5, SGPT], there is a quasi-plurisubharmonic function on $\mathbb P^ n$, not identically $-\infty $, that equals $-\infty $ on $E_ j$. Rescaling it so that it becomes $\theta _1$-plurisubharmonic and subtracting its supremum gives

The $L^1$ norms of these normalized potentials are uniformly bounded by theorem 5.25.2. The partial sums

decrease, have uniformly bounded $L^1$ norms, and remain $\theta _1$-plurisubharmonic because $\theta _1\geq 0$ and the coefficients sum to at most one. By corollary 5.25.3, their limit $p$ belongs to $\mathrm{PSH}(\mathbb P^ n,\theta _1)$. Every point of $E$ is a pole of one of the summands, hence $p=-\infty $ on $E$. By lemma 16.28.9,

has the required properties. Its supremum on $K$ is finite, since $b$ is bounded above on the compact set $K$ and is not identically $-\infty $ on $K$, which is non-pluripolar. Subtracting this supremum gives $b\leq 0$ on $K$. $\square$

Proof. Let $u\in \mathcal L_\delta $ satisfy $u\leq \psi $ on $K\setminus E$, where $E$ is pluripolar. Choose $b$ as in the lemma, with $b=-\infty $ on $E$ and $b\leq 0$ on $K$, and put $m=\inf _ K\psi $. For $\epsilon >0$ define

This function belongs to $\mathcal L_\delta $. On $K\setminus E$, the inequalities $u\leq \psi $, $b\leq 0$ and $m\leq \psi $ give $u_\epsilon \leq \psi $, and on $E$ the pole of $b$ gives $u_\epsilon =-\infty $. Thus $u_\epsilon $ is a competitor for $V_\delta $, so $u_\epsilon \leq F_\delta $; moreover, $u_\epsilon \to u$ almost everywhere as $\epsilon \downarrow 0$. By lemma 5.24.3, $u\leq F_\delta $ everywhere. Taking the supremum over $u$ and regularizing proves the asserted equalities.

Alternatively, apply proposition 5.40.6 on $\mathbb P^ n$ with the big background $\theta _\delta $ and the bounded obstacle $\psi -\delta \chi /2$, and then use (16.28.11). $\square$

Proof. Let $m=\inf _ K\psi $. Inclusion of the growth classes and scaling of competitors give, for $0<\alpha <\beta $,

Indeed, if $u$ is a competitor for $V_\beta $, then $(\alpha /\beta )(u-m)+m$ is a competitor for $V_\alpha $, since $\psi \geq m$. The constant $m$ is also a competitor, so all terms are finite and the right-hand side is nonnegative. The estimate shows that $V_\delta (z)$ depends continuously on $\delta >0$ at every point $z$. Since $V_\delta \leq \psi $ on $K$ and $V_\delta $ increases with $\delta $, the set $\Lambda $ is closed upwards, and by continuity it is relatively closed in $(0,\infty )$. This proves the three possible forms of $\Lambda $.

For $0\leq t\leq 1$, convex combinations of competitors give

By the preceding proposition, the same inequality holds for the regularized envelopes almost everywhere, and hence everywhere by lemma 5.24.3. Thus $F_\delta (z)$ is a concave nondecreasing function of $\delta $ at every point $z$.

For $a\in K$ and $\delta \in \Lambda $, the defect $d_\delta =F_\delta (a)-\psi (a)$ is nonnegative, concave and nondecreasing in $\delta $. Suppose $d_\alpha =0$, and that there is $\eta \in \Lambda $ with $\eta <\alpha $. For every $\beta >\alpha $, write $\alpha $ as a convex combination of $\eta $ and $\beta $ with positive weights. Concavity then forces both nonnegative defects $d_\eta $ and $d_\beta $ to vanish, and monotonicity gives $d_\delta =0$ for all $\delta \in \Lambda $ with $\delta <\alpha $. Hence a zero of the defect at a parameter other than the smallest element of $\Lambda $ forces every defect to vanish. This proves the alternatives for each $\Lambda _{\mathrm{reg}}(a)$, and intersecting them gives the alternatives for $\bigcap _{a\in K}\Lambda _{\mathrm{reg}}(a)$. $\square$

Proof. If $u$ is a competitor for $V_\alpha =V_{\alpha ,\psi }$ and $v$ is a competitor for $V_ K=V_{1,0}$, then $u+(\beta -\alpha )v$ is a competitor for $V_\beta $. Taking the suprema over $u$ and $v$ independently gives

By proposition 16.28.10, the raw envelopes agree almost everywhere with their regularizations, so this inequality holds almost everywhere for the regularizations, and hence everywhere by lemma 5.24.3. This is the first inequality. The same argument applied to (16.28.15) gives the second. $\square$

Proof. For $0<h_1<h_2$, concavity of $F_\delta (a)$ gives

Thus the difference quotients decrease with $h$. Lemma 16.28.18 bounds them between $V_ K^*(a)$ and $(F_\tau (a)-m)/\tau $, which proves the existence and finiteness of $D_\tau (a)$ and (16.28.21). Combining the lower bound for the increments with (16.28.17) gives (16.28.22), with a bound independent of $a$. If $D_\tau (a)=V_ K^*(a)$, then by (16.28.21) every difference quotient equals $V_ K^*(a)$, which gives (16.28.23). Conversely, that formula gives $D_\tau (a)=V_ K^*(a)$. $\square$

Proof. At $\tau $ one has $F_\tau (a)=\psi (a)$, and raw contact holds for all $\delta \geq \tau $ by theorem 16.28.14. If $D_\tau (a)=0$, then by (16.28.21) every difference quotient at $\tau $ vanishes, so $F_\delta (a)=\psi (a)$ for all $\delta >\tau $. If $D_\tau (a)>0$, then $F_\delta (a)>\psi (a)$ for all $\delta $ slightly larger than $\tau $, and by monotonicity for all $\delta >\tau $. In this case $\Lambda _{\mathrm{reg}}(a)$ contains $\tau $ but no larger parameter, so by theorem 16.28.14 it equals $\{ \lambda \} $, where $\lambda =\tau $ is the smallest element of $\Lambda $. $\square$

Proof. On the real Banach space $C(K)$, put $T(f)=F_{1,f}(a)$. By (16.28.25), $T$ is concave, since convex combinations of competitors are competitors. Comparison of obstacles and translation by constants show that $T$ is nondecreasing and satisfies

Fix $f_0\in C(K)$. The directional derivative of the convex Lipschitz function $-T$ at $f_0$ is finite and sublinear. By the Hahn–Banach theorem, there is a continuous linear functional dominated by it, and since difference quotients of a convex function are monotone, this functional supports $-T$ at $f_0$. Changing the sign, we obtain a continuous linear functional $\ell $ with

Testing $f=f_0+th$ with $h\geq 0$ and $t>0$, and using monotonicity, gives $\ell (h)\geq 0$. Testing with $f=f_0\pm t$ gives $\ell (1)=1$. By the Riesz representation theorem, $\ell (f)=\int _ K f\, \mathrm{d}\mu $ for a Borel probability measure $\mu $ on $K$. Set $c=T(f_0)-\int _ K f_0\, \mathrm{d}\mu $. The supporting inequality at $f=0$ gives $c\geq T(0)=V_ K^*(a)\geq 0$. Together with (16.28.25), it shows that $(\mu ,c)\in \mathcal J_ a(K)$ and

Conversely, applying (16.28.26) to $u=F_{1,f_0}$, which lies below $f_0$ quasi-everywhere on $K$, shows that $T(f_0)\leq \int _ Kf_0\, \mathrm{d}\mu +c$ for every pair in $\mathcal J_ a(K)$. This proves (16.28.28) for $\delta =1$, with the minimum attained. The identity $F_{\delta ,q}=\delta F_{1,q/\delta }$ gives the general case, and the zero obstacle gives the last assertion. $\square$

Proof. Each defining inequality of $\mathcal J_ a(K)$ is preserved under weak convergence of the probability measures and convergence of the constants, because the test function $f$ is continuous. Since the Borel probability measures on the compact metric space $K$ form a weakly compact set, the pairs in $\mathcal J_ a(K)$ with $c\leq C$ form a compact set for each finite $C$.

Call a pair in $\mathcal J_ a(K)$ active at $\tau $ if it satisfies the equality in (16.28.30). Such pairs exist by theorem 16.28.27, and since $\int _ K\psi \, \mathrm{d}\mu \geq m=\inf _ K\psi $, they satisfy

The set of active pairs is therefore compact, and the minimum of $c$ over it is attained. By theorem 16.28.27 at $\tau +h$, each active pair gives $F_{\tau +h}(a)\leq F_\tau (a)+hc$, hence $D_\tau (a)\leq c$.

Conversely, for $h>0$ choose a pair $(\mu _ h,c_ h)$ attaining the minimum in (16.28.28) at $\tau +h$, with $q=\psi $. The representation at $\tau $ gives

Thus $0\leq c_ h\leq (F_{\tau +h}(a)-F_\tau (a))/h$, and these constants are bounded by (16.28.21). Along a sequence $h\downarrow 0$, compactness gives a limiting pair $(\mu ,c)\in \mathcal J_ a(K)$. The same bound makes $F_{\tau +h}(a)\to F_\tau (a)$ and $hc_ h\to 0$, and continuity of $\psi $ then gives $\int _ K\psi \, \mathrm{d}\mu +\tau c=F_\tau (a)$. Thus the limit pair is active, and it satisfies $c\leq D_\tau (a)$. This proves (16.28.30). Finally, every pair has $c\geq V_ K^*(a)$ by theorem 16.28.27, so the attainment of the minimum gives the stated criterion. $\square$

Proof. Nonemptiness follows from theorem 16.28.27. Since the test functions are continuous, $\mathcal J_ a(K)$ is closed and convex, and the equality defining active pairs is a closed affine condition. For an active pair, (16.28.17), applied with the weight $q$, gives

The pairs in $\mathcal J_ a(K)$ with bounded constants form a compact set, as in the proof of theorem 16.28.29; hence the set of active pairs is compact.

With $q$ fixed, lemma 16.28.18 and (16.28.17), applied with the weight $q$, bound $|F_{\eta ,q}(a)-F_{\delta ,q}(a)|$ by $|\eta -\delta |(V_ K^*(a)+\operatorname {osc}_ Kq/\epsilon )$. Comparison of obstacles and translation by constants bound $|F_{\eta ,p}(a)-F_{\eta ,q}(a)|$ by $\| p-q\| _\infty $. Adding these bounds gives (16.28.34).

For the sequence in the statement, the constants $c_ j$ have a common bound, by the estimate above, because the weights $q_ j$ are uniformly bounded and the parameters $\delta _ j$ are bounded below by a positive constant. Weak compactness of probability measures gives a convergent subsequence of pairs, and its limit belongs to $\mathcal J_ a(K)$, which is closed. The estimate (16.28.34) gives $F_{\delta _ j,q_ j}(a)\to F_{\delta ,q}(a)$, while uniform convergence of the $q_ j$ and weak convergence of the $\mu _ j$ give $\int _ Kq_ j\, \mathrm{d}\mu _ j\to \int _ Kq\, \mathrm{d}\mu $. Passing to the limit in the equality defining active pairs proves the last assertion. $\square$

Proof. By compactness of the set of active pairs, the minimum on the right is attained. For a fixed pair $(\mu ,c)$ active at $(\delta ,q)$, the representation (16.28.28) at the perturbed parameters gives

Choosing an active pair that attains the minimum on the right, we see that the upper limit of the difference quotients is at most this minimum. Conversely, choose a pair $(\mu _ h,c_ h)$ active at $(\delta +hs,q+hr)$. The representation at $(\delta ,q)$ gives

The estimate (16.28.34) bounds the difference quotients. Along a sequence of values of $h$ realizing their lower limit, lemma 16.28.33 gives a subsequence of pairs converging to a pair active at $(\delta ,q)$. For this pair, $\int _ Kr\, \mathrm{d}\mu +sc$ is at least the minimum on the right. This proves (16.28.36).

For convergent directions, the same two inequalities hold. The fixed active pair attaining the minimum gives an upper bound with the right limit. The active pairs at the perturbed parameters still have uniformly bounded constants, and their subsequential limits are active at $(\delta ,q)$. Uniform convergence of the $r_ j$ controls the integrals $\int _ Kr_ j\, \mathrm{d}\mu _{h_ j}$, and $s_ j\to s$ controls the terms $s_ jc_{h_ j}$. The lower bound therefore has the same limit. Note that all perturbed growth allowances $\delta +h_ js_ j$ are positive for small $h_ j$. $\square$

Proof. Apply (16.28.36) with $r=0$ and $s=\pm 1$. The direction $s=-1$ gives minus the left derivative, which proves (16.28.38). Concavity gives

Hence the open intervals $(D^+_{\delta ,q}(a),D^-_{\delta ,q}(a))$, which are nonempty exactly at the parameters where differentiability fails, are pairwise disjoint. Choosing a rational number in each of them proves countability. $\square$

Proof. With $s=0$, (16.28.36) shows that the right derivative in the direction $r$ is the minimum of $\int _ Kr\, \mathrm{d}\mu $ over active pairs, and the direction $-r$ shows that the left derivative is the maximum. If all active pairs have the same probability measure, these values agree and define a continuous linear derivative. Conversely, if the map is Gâteaux differentiable, the minimum and the maximum agree for every continuous $r$. Continuous functions separate Borel probability measures on $K$, so all active pairs have the same probability measure; their constant is then determined by (16.28.32), since $\delta >0$. $\square$

Proof. For $t\geq 0$, each competitor is at most $-t$ on the unit circle, so the maximum principle for subharmonic functions bounds it by $-t$ on $D$. This bound persists after regularization on $D$, and the constant $-t$ is a competitor. Hence $F_{\delta ,q_ t}(0)=-t$.

For $t=-b$ with $0\leq b\leq \delta /2$, define

This function is continuous and subharmonic: its Laplacian has density $4b$ inside the disk, density zero outside, and a nonnegative contribution on the circle coming from the jump $\delta -2b\geq 0$ of the radial derivative. It belongs to $\mathcal L_\delta $ and equals $q_{-b}$ on $K$. Hence $F_{\delta ,q_{-b}}(0)\geq u_ b(0)=0$. On $D$ the raw envelope is at most the continuous function $b|z|^2$, so its regularization at zero is at most zero. This proves (16.28.41). In particular, the right and left derivatives in $t$ at $t=0$ are $-1$ and $0$, and corollary 16.28.39, applied in the direction $r=-|z|^2$, gives the final assertion. $\square$

Proof. In the pluripolar case, choose $b\in \mathcal L_1$ with $b=-\infty $ on $E$ by lemma 16.28.12, and a finite upper bound $C$ for $b$ on $B$. The functions

are competitors for $h_{E,B}$, and they tend to zero almost everywhere as $\epsilon \downarrow 0$. Hence $h_{E,B}^*\geq 0$ almost everywhere, and then everywhere by lemma 5.24.3. Since $h_{E,B}^*\leq 0$, it vanishes.

Now suppose that $E$ is non-pluripolar. Proposition 16.28.10 gives a finite upper bound $M\geq 1$ for $V_ E^*$ on $B$. For each competitor $v$ for $V_ E$, the function $\max (-1,v/M-1)$ is a competitor for $h_{E,B}$. Taking suprema and regularizing gives

As $V_ E^*\geq 0$, the equality $h_{E,B}^*(a)=-1$ forces $V_ E^*(a)=0$.

For the converse, choose $r<R_0<R$ and $k>0$ with $k\log (R_0/r)>1$. The plurisubharmonic function

belongs to $\mathcal L_ k$ and equals $-\infty $ at $a$. For a competitor $u$ for $h_{E,B}$, define $W=\max (u+1,L)$ on $B$ and $W=L$ outside $B$. On the collar $R_0<\lVert z-a\rVert <R$ we have $L>1\geq u+1$, so the two definitions agree there, and $W$ is plurisubharmonic by lemma 5.24.2. On $E$ one has $u=-1$ and $L\leq 0$, so $W=0$; on the ball $B(a,r)$ one has $L<0\leq u+1$, so $W=u+1$. Hence $W/k$ is a competitor for $V_ E$, with $k$ independent of $u$. Taking the supremum over $u$ and regularizing at $a$ gives

Thus $V_ E^*(a)=0$ forces $h_{E,B}^*(a)=-1$. $\square$

Proof. Put $M_ r=\sup _{K_ r}q$. Raw contact, monotonicity of envelopes with respect to the compact set, and (16.28.17) for $K_ r$ give

Since $q$ is upper semicontinuous and $a\in K_ r$, we have $M_ r\downarrow q(a)$ as $r\downarrow 0$, which gives $F_{\delta ,K,q}(a)=q(a)$.

Every compact subset $E\subseteq K$ inherits raw contact: on $E$, the raw envelope for $E$ is at least the raw envelope for $K$, which equals $q$, and at most the obstacle $q$. For fixed $s>0$, apply the same bounds with $r<s$ and $K_ s$ in place of $K$, and let $r\downarrow 0$. This proves local $(\delta ,q)$-regularity. By theorem 16.28.14, raw contact also holds for every larger parameter, and applying what has been proved to each of them gives the last assertion. $\square$

Proof. Set $B=B(a,R)$ and $E=K_ r$. For a competitor $u$ for $h_{E,B}$, the function

is plurisubharmonic on $B$, because the assumption on $\mathrm{dd}^{\mathrm{c}}\Psi $ makes $\Psi -c\lVert z-a\rVert ^2$ plurisubharmonic there. Since $u\leq 0$, we have $H_ u\leq \Psi $ where $\lVert z-a\rVert \geq r$. Define $W=\max (\Psi ,H_ u)$ on $B$ and $W=\Psi $ outside $B$. The two definitions agree near $\partial B$, so $W$ is plurisubharmonic on $\mathbb C^ n$, and it belongs to $\mathcal L_\tau $ because it equals $\Psi $ outside $B$. On $E$, the equality $u=-1$ gives $H_ u\leq \Psi $, and on $K\setminus E$ the distance bound gives the same inequality. Thus $W=q$ on $K$, so $W$ is a competitor for $F_{\delta ,K,q}$ for each $\delta \geq \tau $. Since $W\geq H_ u$ near $a$, taking the supremum over $u$, regularizing, and using the continuity of $\Psi $, we obtain (16.28.47).

The function $\Psi $ itself is a competitor, so raw contact holds for every $\delta \geq \tau $. If regular contact holds for one such $\delta $, the left-hand side of the estimate vanishes, so $h_{K_ r,B(a,R)}^*(a)=-1$ for all $0<r<R$. By lemma 16.28.43, every $K_ r$ with $r<R$ is then non-pluripolar and satisfies $V_{K_ r}^*(a)=0$; for $r\geq R$ this follows by monotonicity. Conversely, local pluriregularity and lemma 16.28.45 give regular contact at $a$ for every $\delta \geq \tau $. $\square$

Proof. The complex Hessian of $\Psi $ is semipositive. If its top power did not vanish at $a$, it would be positive definite at $a$, and by continuity $\mathrm{dd}^{\mathrm{c}}\Psi \geq c\omega _0$ on a neighborhood of $a$ for some $c>0$. Theorem 16.28.46 would then give regular contact at $a$ for every $\delta \geq \tau $, a contradiction. $\square$

Proof. Since $r_ j<2^{-j-3}$, the small disks are pairwise disjoint and disjoint from $\overline D$, and they accumulate only at $2$. This proves compactness. Every neighborhood of $2$ contains one of the small disks, and every other point of $E$ lies in the closure of the interior of one of the disks.

To show that the series defines a subharmonic function, write $\chi (\zeta )=\log (1+|\zeta |^2)$ on $\mathbb C$ and $\theta =\omega _{\mathrm{FS}}/2$ on $\mathbb P^1$. By lemma 16.28.9, each

extends to a $\theta $-plurisubharmonic function on $\mathbb P^1$. The Cauchy–Schwarz inequality gives $q_ j\leq \log 4$, while the value of $q_ j$ at infinity is zero. Thus $0\leq \sup q_ j\leq \log 4$, and theorem 5.25.2 bounds $\| q_ j-\log 4\| _{L^1}$ uniformly. The partial sums

decrease, remain $\theta $-plurisubharmonic and have uniformly bounded $L^1$ norms. By corollary 5.25.3, their limit $w$ belongs to $\mathrm{PSH}(\mathbb P^1,\theta )$. Since $\sum _ j2^{-j}=1$,

on the affine chart. This proves subharmonicity and the growth bound, and $\sum _ j j2^{-j}=2$ gives $h(2)=0$.

On the $j$th small disk, all distances to the other centers are less than one, so the corresponding terms are negative, and

On $\overline D$ all distances to the centers are less than four, hence $h\leq 4\log 2$ and $v\leq 0$. On each small disk $|\zeta |<4$, so $9v\leq 14\log 2-16<0$. The growth bound for $v$ and its value at $2$ follow directly from the formula. $\square$

Proof. By the preceding lemma, every neighborhood of a point of $K$ contains a nonempty open subset of $K$, namely a product of open disks. This proves the first assertion, which applies in particular to each $K\cap \overline B(a,r)$, $r>0$.

Put $b(\zeta )=\log |\zeta -2|-\log 3$. Since $|\zeta -2|\leq 3$ on $E$, we have $b\leq 0$ there, and $b(2)=-\infty $. For $\varepsilon >0$, the plurisubharmonic function

belongs to $\mathcal L_1$ and satisfies $u_\varepsilon \leq 0$ at every point of $K$, including the points with $z_1=2$, where it equals $-\infty $. Thus $u_\varepsilon \leq F_{1,K}$. Letting $\varepsilon \downarrow 0$ gives $v(z_1)\leq F_{1,K}(z)$ off the hyperplane $\{ z_1=2\} $, and lemma 5.24.3 extends this inequality everywhere. The value $v(2)=4\log 2/9$ gives the lower bound for $\delta =1$. Since $0$ is a competitor, $V_{\delta ,K}|_ K=0$, and scaling of competitors gives $F_{\delta ,K}=\delta F_{1,K}$. Replacing $K$ by a smaller compact set only increases the envelope, which proves the local assertion.

Finally, let $f$ be plurisubharmonic near $a_0$. If $f(a_0)=-\infty $, (16.28.53) is automatic. Otherwise, restrict $f$ to the complex line through $a_0$ in which only the second coordinate varies. Small circles in this line centered at $a_0$ lie in $K\setminus \{ a_0\} $, and the submean inequality on these circles gives (16.28.53). $\square$

Proof. Choose $\rho \in \mathrm{PSH}(X,\theta )$ and $a>0$ with $\theta _\rho \geq a\omega $, normalize $\rho \leq 0$, and put

The obstacle is finite almost everywhere and bounded below, so lemma 5.40.2 shows that $p$ is bounded. The obstacle is quasi-continuous off the pluripolar pole set of $\rho $. The contact theorem, theorem 16.28.5, applied in the Hermitian background $a\omega $, shows that $\mathrm{MA}_{a\omega }(p)$ is carried by $D=\{ p+\rho =u\} $, outside the pole set of $\rho $.

By lemma 5.24.3, $p+\rho \leq u$ everywhere, and

On common finite truncation levels, positivity of the mixed products and ordered contact give

Exhausting by these levels proves the same inequalities for the extended measures. Since $p$ is bounded, its Monge–Ampère mass with respect to the Hermitian form $a\omega $ is positive by corollary 14.19.11. Integrating the inequalities proves the assertion. $\square$

Proof. If the conclusion failed, we could choose $a>0$ such that $E=\{ u<v-a\} $ is nonempty. This set has positive volume by lemma 5.24.4. Fix $C$ with $|u|,|v|\leq C$.

For $b>1$ define the bounded envelope

Since $\theta \geq 0$, constants below the obstacle are competitors. By the bounded contact theorem, theorem 16.28.4, the measure $\mathrm{MA}_\theta (w_ b)$ is concentrated on $D_ b$. The potential $b^{-1}w_ b+(1-b^{-1})v$ lies below $u$ quasi-everywhere, hence everywhere by lemma 5.24.3, and it equals $u$ on $D_ b$ outside a pluripolar set. Ordered contact and positivity of the mixed terms give

Choose $b$ so large that $(1-b^{-1})^ n>c$. Then the hypothesis forces $\mathrm{MA}_\theta (w_ b)=0$ on $D_ b\cap \{ u<v\} $. Since $\theta $ is non-collapsing, $w_ b$ has positive total mass, so some of this mass is carried by $D_ b\cap \{ u\geq v\} $. At a point of this set outside the exceptional pluripolar set,

Consequently $\sup _ Xw_ b\geq -C$ for all large $b$.

On the other hand, on the fixed set $E$ the obstacle bound gives $w_ b\leq C-(b-1)a$ quasi-everywhere. Hence

for all sufficiently large $b$. This contradicts the uniform $L^1$ bound for normalized potentials in theorem 5.25.2. $\square$

Proof. Choose a proper analytic set $S\subset X$ such that $\mu $ is an isomorphism outside $D=\mu ^{-1}(S)$. On these complements, bounded truncations commute with composition by $\mu $, and their products commute with the biholomorphic change of coordinates. Passing to the increasing limits gives the measure identity there. By theorem 13.14.3, the two measures give no mass to $S$ and $D$, which proves the global identity and equality of total masses.

Pullback is a bijection between the potentials in the two backgrounds by proposition 5.43.18. Surjectivity of $\mu $ preserves and reflects bounded differences, so this bijection identifies the ordinary-type competitor families. Their masses agree by the first assertion. Taking their infima and suprema proves the volume identities. $\square$

Proof. If $u\geq \varphi -C$, then $h=\varphi -bC$ belongs to $\mathcal A_ b(u,\varphi )$ for each $b$; hence every $u\sim \varphi $ belongs to the class. More generally, if $u'\geq u-C$ and $h\in \mathcal A_ b(u,\varphi )$, then $h-bC\in \mathcal A_ b(u',\varphi )$. Since the condition $u\preceq \varphi $ is also unchanged by bounded changes of $u$, the class is stable under such changes. A convex combination of two competitors is a competitor for the same convex combination of the two potentials, which proves convexity. Finally, a competitor for $u$ is a competitor for any larger $v$, and the assumption $v\preceq \varphi $ gives the remaining condition.

If $\varphi '\in \mathrm{PSH}(X,\theta )$ and $\varphi '\sim \varphi $, choose $C\geq 0$ with $|\varphi '-\varphi |\leq C$ off their common pole set. For $h\in \mathcal A_ b(u,\varphi )$, the function $h-(b-1)C$ belongs to $\mathcal A_ b(u,\varphi ')$: the required inequality holds off the poles, and it extends everywhere by lemma 5.24.3. The reverse implication follows by interchanging the references, and $u\preceq \varphi $ is equivalent to $u\preceq \varphi '$.

For the last assertion, note that $h\mapsto h-f$ is a bijection from $\mathrm{PSH}(X,\theta )$ onto $\mathrm{PSH}(X,\theta +\mathrm{dd}^{\mathrm{c}}f)$. It preserves the order $\preceq $ and transforms each stretching inequality into

This identity also holds at poles, because $f$ is smooth and finite. Thus the corresponding families of competitors are in bijection for every $b\geq 1$. $\square$

Proof. Fix $b,s\geq 1$ and put $h=Q_{bs}(u,\varphi )$. Its defining inequality gives, off the pole sets,

The left-hand side is $\theta $-plurisubharmonic, so it is at most $Q_ b(u,\varphi )$. Hence $h\in \mathcal A_ s(Q_ b(u,\varphi ),\varphi )$ for every $s$. Also $Q_ b(u,\varphi )\leq u\leq \varphi $, so $Q_ b(u,\varphi )\preceq \varphi $. The inequalities between convex combinations extend across the poles by lemma 5.24.3, which proves the first assertion.

For the second assertion, choose a constant $C$ such that $u-C\leq \varphi $ and $w-C\leq \varphi $. Since $Q_ b(u-C,w-C)=Q_ b(u,w)-C$ and class membership is unchanged by translation, we may assume $u,w\leq \varphi $. Because $u\leq w\leq \varphi $, the obstacle for $Q_ b(u,w)$ lies above that for $Q_ b(u,\varphi )$ and below $u$. Thus

Since $Q_ b(u,\varphi )\in \mathcal E_\varphi $ by the first assertion, the stability under increase proved in proposition 16.28.62 gives the conclusion. $\square$

Proof. Adding a common constant to all $u_ i$, we may assume $u_ i\leq \varphi $. By lemma 16.28.63, proposition 16.28.62,

Each summand is at most $\varphi $. For every $i$, the defining inequality of $Q_ N(u_ i,\varphi )$ therefore gives

These inequalities between convex combinations also hold at common poles. Thus $h$ is a competitor for $p=P_\theta (\min _ i u_ i)$, so $p$ is not identically $-\infty $, and $h\leq p\leq u_1\preceq \varphi $. Stability under increase gives $p\in \mathcal E_\varphi $, and undoing the translation preserves this conclusion. $\square$

Proof. The relation $u\preceq \varphi $ gives $P_{\theta ,F}[u]\leq P_{\theta ,F}[\varphi ]$. For the reverse inequality, let $a\in (0,1)$ and choose $h_ a\in \mathcal A_{1/(1-a)}(u,\varphi )$, so that $a\varphi +(1-a)h_ a\leq u$. Combining competitors below $F$ gives

Indeed, if $g_1$ and $g_2$ are competitors for $P_{\theta ,F}[\varphi ]$ and $P_{\theta ,F}[h_ a]$, then $ag_1+(1-a)g_2$ lies below $F$ and is at least as singular as $a\varphi +(1-a)h_ a\leq u$; taking the two upper envelopes preserves the inequality, by local regularization. Write $m=\min _ XF$ and $M=\max _ XF$. Translating competitors gives

Every $P_\theta [h_ a]$ has supremum zero, since it lies above $h_ a-\sup _ Xh_ a$ and below zero. The uniform $L^1$ bound for normalized potentials in theorem 5.25.2 therefore bounds the $L^1$ norms of the $P_{\theta ,F}[h_ a]$ uniformly in $a$. Letting $a\uparrow 1$ in the first inequality gives $P_{\theta ,F}[\varphi ]\leq P_{\theta ,F}[u]$ almost everywhere, and by lemma 5.24.3 the inequality holds everywhere. $\square$

Proof. Since $\varphi \leq 0$, the potential $u$ lies below the obstacle $u-\varphi $ away from the pole set of $\varphi $, so it is a competitor. Hence the envelope $w$ is not identically $-\infty $. It is at most zero, since $u-\varphi \leq 0$, and it satisfies $w+\varphi \leq u$ off the poles, hence everywhere by lemma 5.24.3. For $b\geq 1$, let $h_ b=Q_ b(u,\varphi )$. Then, off the poles,

Since $\theta \geq 0$, the function $b^{-1}h_ b$ is $\theta $-plurisubharmonic, and so it is a competitor for $w$. Hence $h_ b\leq bw$, so $\mathcal A_ b(w,0)$ is nonempty for every $b$. Since also $w\leq 0$, this means exactly that $w\in \mathcal E_0$. $\square$

Proof. The obstacle $u-q$ is nonnegative away from the poles, so by lemma 5.40.2 the envelope $p=P_{(1-a)\theta }(u-q)$ is bounded. By the contact theorem, its measure is carried by $D=\{ p+q=u\} $, and $p+q\leq u$ everywhere. Since $p+q$ is $\theta $-plurisubharmonic, positivity of mixed products and ordered contact imply

where all assertions are first restricted to the complement of the poles. Writing $p=(1-a)\widetilde p$ with $\widetilde p\in \mathrm{PSH}(X,\theta )\cap L^\infty (X)$ and integrating, we obtain the lower bound from the definition of $v_-(\theta )$. Finally, the bounded potential $p$ has strictly positive mass by proposition 16.28.55, since $(1-a)\theta $ is big. $\square$

Proof. After adding a constant to $u$, we may assume $u\leq \varphi $. Let $w\in \mathcal E_0$ be as in lemma 16.28.68. Since $w\in \mathcal E_0$, the envelope $P_\theta (w/\delta )$ is not identically $-\infty $, and $\delta P_\theta (w/\delta )\leq w$. The potential

belongs to $\mathrm{PSH}(X,(\varepsilon +\delta )\theta )$, so it is a competitor for $P_{(\varepsilon +\delta )\theta }(u)$. Applying lemma 16.28.69 with $a=\varepsilon +\delta $ and letting $\delta \downarrow 0$ gives the uniform estimate. The positivity follows from the same lemma with any one such $\delta $. $\square$

Proof. Use the smooth nonpositive $\omega $-plurisubharmonic functions $u_ j$, $j\geq 1$, of theorem 13.14.18 and their convergent series $U=\sum _{j\geq 1}2^{-j}u_ j$. For each integer $N\geq 1$, write

The head $A_ N$ is smooth, and since the finite partial sums of the tail decrease, $V_ N\in \mathrm{PSH}(X,\delta _ N\omega )$; the tail is not identically $-\infty $, because $U=A_ N+V_ N$ is not.

Fix $b>1$ and choose $N$ so large that $b\delta _ N\leq 1$. Then

is $\omega $-plurisubharmonic, because $\mathrm{dd}^{\mathrm{c}}h_ b\geq -b\delta _ N\omega \geq -\omega $. Moreover, $h_ b\leq bU$. Thus $h_ b\in \mathcal A_ b(U,0)$ for every $b$, and since $U\leq 0$, this proves $U\in \mathcal E_{\mathrm{env}}(X,\omega ,0)$. For $\varepsilon U$, use the tails $\varepsilon V_ N$, and choose $N$ with $b\varepsilon \delta _ N\leq 1$. By proposition 16.28.65, the singularity envelope of each of these potentials is $P_\omega [0]=0$. Their infinite mass and trivial multiplier ideals are given by corollary 13.14.19. $\square$

Proof. If $v_+(\omega )<+\infty $, all $\omega $-plurisubharmonic functions have finite extended mass by proposition 13.14.16. If $v_+(\omega )=+\infty $, proposition 16.28.72 contradicts each of the other two conditions. $\square$